Bunuel
In the figure above, two security lights, L1 and L2 , are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chain-link fence. What is the total length s of fence, in feet, illuminated by the two lights?
(A) 260
(B) 240
(C) 220
(D) 200
(E) 180
Attachment:
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Solution:Attachment:
circles Overlapping1.png [ 30.55 KiB | Viewed 2103 times ]
Let’s name the 4 intersection points of the fence with the circles as A, B, C, and D (see diagram above) and notice that we need to find the length of AD. Furthermore, let’s name one more point, E, where E is the midpoint of AC. We see that AEL1 is a right triangle; in fact it’s a 3-4-5 right triangle. Therefore, AE = 80 and AC = 2 x 80 = 160. Since BD = AC,then BD = 160 also. Notice that AD = AC + BD - BC, AD = 160 + 160 - BC = 320 - BC. If we can find the length of BC, we can determine the length of AD. However, we can see that BC has a length of less than 80 since it is less than EC, which has a length of exactly 80. Since BC < 80, therefore AD = 320 - BC will be greater than 240. Since the only answer choice that is greater than 240 is 260, choice A must be the correct answer.
(Note: This means BC must be 60, but we will leave the readers to verify this on their own.)
Alternate Solution:Attachment:
Circles Overlapping2.png [ 28.09 KiB | Viewed 2093 times ]
Note that the distance S is equal to the distance from A to D. We can find the length of AB, the length of CD, and then the length of BC.
First recognize triangle ABL1 as a 3-4-5 triangle, with lengths 60, 80, and 100. Thus, the length AB = 80. This is the same as the length CD because triangle DCL2 is also a 3-4-5 triangle, so CD = 80.
Note that the distance from B to C is equal to the distance from L1 to L2, and this is given as 100.
Thus, the total distance is 80 + 100 + 80 = 260.
Answer: A